Chapter 8
Predicting What Comes Next: Exploring Sequences and Progressions
CBSE Class 9
Mathematics Solutions
Educart Mathematics class Class 9 NCERT Exemplar cover
Question:

Which term of the G.P.: 2, 6, 18, ...  is 4374? Write the explicit formula as well as the recursive formula for the nthn^{\text{th}} term.                                                                                                                                                                                        [Page No. 193]

Answer: Verified

Here, a=2,r=3\text{Here, }a=2,r=3

General term: Tn=2(3n−1)T_n = 2(3^{n-1})

∴2(3n−1)=4374\therefore 2(3^{n-1}) = 4374

3n−1=21873^{n-1} = 2187

2187=372187 = 3^7

So, n−1=7\text{So, } n - 1 = 7

n=8n = 8

Thus, 8th term of the G.P. is 4374.

Recursive formula: T1=2,Tn=3Tn−1T_1 = 2, T_n = 3T_{n-1} for (n≥2)(n \geq 2).

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More NCERT Questions
Can the same approach (refer to NCERT page no. 183) be used to find the sum of 1 + 2 + 3 + + 100? [Page No. 184] Observe the Sierpiński triangle. (A) How many black triangles are there in Stages 0 to 3? (B) Predict the number at Stages 4 and 5. (C) Find a rule for the number of black triangles at the nth stage. Can you use this formula to find S20, S50 or S1000? [Page No. 185] Let us revisit the sequence tn of triangular numbers 1, 3, 6, 10, 15, shown in the given figure. Note that the nth term of this sequence is the sum of the first n natural numbers. Thus, tn = n(n + 1)/